Jul 14 Cardinality

Size of Sets

Cardinality is defined as the number of elements in a set.

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The set $\{1,2,3,4,5\}$ has cardinality 5. What about the set $\{a,b,c,d,e\}$? Well, we can map each element in the first set to an element in the second set
$$\begin{cases}1 \to a\\2 \to b \\ 3 \to c \\ 4 \to d \\5 \to e\end{cases}$$
If we can do this, then the two sets have the same cardinality. What if the sets are infinite? Consider the natural numbers below:
$$\{0,1,2,3,4,5\dots\}$$
The cardinality of that set is called $\aleph_0$ (Aleph null). This is called countable infinity, because every element can be counted one at a time, and each element is associated with a natural number.
What about $\{0, 2, 4, 6, 8, 10 \dots\}$? What's the cardinality of that set? It turns out, it's still $\aleph_0$. Let's try to map them.
$$\begin{cases}0 \to 0\\1 \to 2 \\ 2 \to 4 \\ 3 \to 6 \\4 \to 8\end{cases}$$
So, any set that can be mapped to the naturals is countable. Let's explore the uncountables.

Uncountable

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Is there anything that cannot be counted? Surely there must be some set of numbers that is a magnitude greater than the natural numbers. There is! Let's look at all possible decimals from 0 to 1. We will represent them as binary numbers beginning with the digit after the decimal.

Proof

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First, let's re-establish what we're trying to prove. We're trying to show that this set X isn't in $S_1$ to $S_n$.
Second, let's re-establish what the set is. The set is the set such that nth element is not in $S_n$.
Now, we do a proof by contradiction. We say this set X equals $S_i$ for some $i$. This means that the i'th element is the i'th element of $S_i$. However, we know this is false. X is the set $X_i s.t. X_i \neq S_{ii}$ Therefore, X will differ at some point with some binary number. So, X doesn't equal an arbitrary $S_i$, and X is not in the set of all binary numbers.
This means that there is a binary number that was never counted. Therefore, the decimals from 0 to 1 are uncountable.

Cardinality of the Decimals

As we showed above, decimals could be represented by a set of 1s and 0s.

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As we showed above, decimals could be represented by a set of 1s and 0s. Each position has 2 options: 0 or 1, and there are $\aleph_0$. Therefore, the number of decimals is $$2 \times 2 \times \dots \left(\aleph_0 \text{ times}\right) = 2^{\aleph_0}$$

Continuum Hypothesis

In 1900, Hilbert published a list of 23 problems that he wanted to either prove or disprove. One of the more famous problems is the continuum hypothesis which states there is no set with a cardinality between that of the integers and the real numbers.